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m_a_m_a [10]
3 years ago
9

In a collection of nickels, dimes, and quarters worth $6.90, the ratio of the number of nickels to dimes is 3:8. The ratio of th

e number of dimes to quarters is 4:5. Find the number of each type of coin.
Mathematics
1 answer:
dimulka [17.4K]3 years ago
8 0
N=number of nickels
d=number of dimes
q=number of quarters

example
if n:d=3:8
lets say n=3 and d=8
3*8=8*3
so
8n=3d
slightly confusing but go with it, it migh make sense or it might not


n:d=3:8
8n=3d

d:q=4:5
5d=4q

solve for one of them
solve for nickels in first and quarters in second

8n=3d, n=(3/8)d
5d=4q, (4/5)d=q

5n+10d+25q=690
divide by 5 to make it easeir
n+2d+4q=138
sub (3/8)d for n
sub (4/5)d for q
(3/8)d+2d+5(5/4)d=138
(3/8)d+2d+25/4d=138

3/8+2+25/4=
3/8+16/2+50/8=
69/8

(69/8)d=138
times both sides by 8
69d=1104
divide both sides by 69
d=16

16 dimes
sub back

(5/4)d=q
(5/4)16=q
80/4=q
20=q

(3/8)d=n
(3/8)16=n
48/8=n
6=n



6 nickles
16 dimes
20 quarters
yah
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QUESTION A

The given multiplication problem is

\frac{39}{64} \times \frac{8}{13}

Factor each term to obtain;

\frac{13\times 3}{8\times8} \times \frac{8}{13}

Cancel out the common factors to obtain;

\frac{1\times 3}{8\times1} \times \frac{1}{1}

Simplify to get;

\frac{3}{8}

QUESTION B

The given multiplication problem is

\frac{2}{3}\times \frac{1}{5}\times \frac{4}{7}

This the same as

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The given problem is

\frac{3}{5}\times \frac{10}{12} \times \frac{1}{2}

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QUESTION D.

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\frac{4}{9}\times 54

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\frac{32}{3} \times \frac{11}{8}

Cancel common factors to get;

=\frac{4}{3} \times \frac{11}{1}

Simplify

=\frac{44}{3}

Convert back to mixed numbers;

=14\frac{2}{3}

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